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Find the characteristic frequencies of a circular membrane which satisfies the Klein Gordon equation (Problem 25).

Short Answer

Expert verified

The characteristic frequencies of oscillation isν=v2πknma2+λ2 .

Step by step solution

01

Given information:

Klein-Gordon equation is 2u=1v22ut2+λ2u.

02

Definition of Wave:

Any disturbance or energy transfer from one location to another is referred to as a wave. Each wave is guided by a mathematical formula. A wave can be either standing or stationary.

03

Write the Klein Gordon equation in polar coordinates:

Consider the Klein Gordon equation.

2u=1v22ut2+λ2u

Express it in polar coordinates.

2u=2ur2+1rur+1r22uθ2 ….. (1)

Separate the spatial and temporal variables to write the above equation in the form of u=Fr,θTt.

Assume the separation constant.-k2

2Fr2+1rFr+1r22Fθ2+k2-λ2F=0 ….. (2)

d2Tdt2+k2v2T=0 ….. (3)

Solve equation (3) to get T=CoskvtSinkvt.

Separate spatial variable from Fr,θand express in terms of Fr,θ=RrΘθ.

1Θ2Θθ2=-n2

Solve the above equation to getΘ=cosnθsinnθ .

Write equation (2) for r.

r22Rr2+rRr-m2R+k2-λ2r2R=0

Solve the equation by changing the variables z=k2-λ212r.

This gives us the Bessel’s differential equation and the solution can be written in the form of R=Zpz.

Change to the previous variable.

R=Zpk2-λ2r

04

Apply Boundary conditions:

Consider the requirement of finite u at r=0.

The general solution isu=Jpk2-λ2rcosnθsinnθcoskvtsinkvt .

Consider the requirement of u=0.

At r=R the equation is:

k2-λ2R=knm ….. (4)

Square both sides in equation (4) and solve for k2.

k2=knmR2+λ2 ….. (5)

As Jpknm=0, knmis a solution of Jpz.

Replace R=a in equation (5) and find the characteristic frequencies of oscillation.

ν=ω2π=kv2π=v2πknma2+λ2

Hence the characteristic frequencies of oscillation is v=v2πknma2+λ2.

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